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User blog:Bubby3/Taranovsky's C 1st system introduction and rules.
Taranovsky's ordinal notation (C(a,b)) might seem complex at the surface, but his 1st system is rather simple if you take an alternate way of explaining it. However, this function gets really complex and weird at 2nd system and higher, which I won't explain in this blog post. My explanation of this ordinal function. Taranovsky's ordinal function is not your typical ordianl collapsing function. First of all, numbers (other than 0), operators, etc., are not defined in this system. Instead, you have to use this ordinal notation to define them with his notation. So when I am saying an expression is equlivant to an ordinal, I am saying that that the ordinal is shorthand for that expression for now and no, to save space. Another weird thing about it which seperates it from most other ordinal collapsing functions is that is evaluated from out to in. So in C(a,b) when b is a limit ordinal, you do not take the fundemental sequence of b. Instead, b has a priority lower than a does. Lets get going with this notation. This part will only help you get intution of this function. The formal definiton will be in next section. C(0,0) is equlivant to 1 C(0,n) is equlivant to n+1. This is even true for infinite or uncountable n. C(1,0) is the limit of C(0,0), C(0,C(0,0)), C(0,C(0,C(0,0))), etc., which means it is equivlant to \(\omega\) C(1,n) is the limit of C(0,n), C(0,C(0,n)), C(0,C(0,C(0,n))), etc., which means C(1,n) is equlivant to \(n + \omega\). That means C(1,C(1,0)) ie equlivant to \(\omega \cdot 2\) In general, C(m+1,n) is the limit of C(m,n), C(m,C(m,n)), C(m,C(m,C(m,n))), etc. When the first element is a limit (cannot be expressed in the form C(0,a) for some a, we take the fundemental sequence of that ordinal, so for example \(C(\omega,0)\) is the limit of C(1,0), C(2,0), C(3,0), etc. Uncountable ordinals \(C(\Omega,0)\) is equal to the first fixed-point of a -> C(a,0). And we can apply any operator we did before to \(C(\Omega,0)\). \(C(\Omega,0)\) is equlivant to \(\varepsilon_0\) \(C(0,\varepsilon_0)\) is equlivant to \(\varepsilon_0 + 1\) \(C(1,\varepsilon_0)\) is equlivant to \(\varepsilon_0 + \omega\) \(C(\varepsilon_0,\varepsilon_0)\) is equlivant to \(\varepsilon_0 \cdot 2\) \(C(\varepsilon_0 + 1,\varepsilon_0)\) is equlivant to \(\omega^{\varepsilon_0 + 1}\) \(C(n,\varepsilon_0)\) is equlivant to \(\omega^{n}\) for \(n > \varepsilon_0\) \(C(\Omega,\varepsilon_0)\) is equlivant to \(\varepsilon_1\) In general, \(C(\Omega,n)\) is the limit of C(0,n), C(C(0,n),n), C(C(C(0,n),n),n), etc. \(C(\Omega,\varepsilon_1)\) is equlivant to \(\varepsilon_2\) \(C(\Omega + 1,0)\) is equlivant to \(\varepsilon_\omega\) \(C(\Omega \cdot 2,0)\) is equlivant to \(\psi(\Omega)\) \(C(\Omega \cdot 3,0)\) is equlivant to \(\psi(\Omega^2)\) \(C(\Omega^2,0)\) is equlivant to \(\psi(\Omega^\Omega)\) \(C(\Omega^2+\Omega,0)\) is equlivant to \(\psi(\Omega^{\Omega+1})\) \(C(\Omega^2 \cdot 2,0)\) is equlivant to \(\psi(\Omega^{\Omega \cdot 2})\) \(C(\Omega^3,0)\) is equlivant to \(\psi(\Omega^{\Omega^2})\) \(C(\Omega^4,0)\) is equlivant to \(\psi(\Omega^{\Omega^3})\) \(C(\Omega^\Omega,0)\) is equlivant to \(\psi(\Omega^{\Omega^\Omega})\) Etc. Here are the uncountable shorthands I used. \(C(0,\Omega)\) is equlivant to \(\Omega + 1\) \(C(1,\Omega)\) is equlivant to \(\Omega + \omega\) \(C(\Omega,\Omega)\) is equlivant to \(\Omega \cdot 2\) \(C(\Omega,C(\Omega,\Omega))\) is equlivant to \(\Omega \cdot 3\) \(C(C(0,\Omega),\Omega)\) is equlivant to \(\Omega \cdot \omega\) \(C(C(\Omega,\Omega),\Omega)\) is equlivant to \(\Omega^2\) \(C(C(\Omega,C(\Omega,\Omega)),\Omega)\) is equlivant to \(\Omega^3\) \(C(C(C(0,\Omega),\Omega),\Omega)\) is equlivant to \(\Omega^\omega\) \(C(C(C(\Omega,\Omega),\Omega),\Omega)\) is equlivant to \(\Omega^\Omega\) The rules Definition of countable: C(a,b) is countable iff b is countable, and 0 is countable. First, the pointer is before the outermost C. The pointer is a > symbol #If the pointer is is before a C, the pointer will jump in the C, and keep repeaking this process #If the pointer is at a 0: ##If the expression is on the outermost layer (or not after a C), C(0,a) is the sucessor of a. The process ends ##If the expression is contained in another C, C(C(0,a),b)n C(a,C(a...C(a,b)) with n 'C(a,'s. The process ends #If the pointer is at an \(\Omega\), then find the innermost countable expression contained in that \(\Omega\), call that A. Make X and Y so that A = X>\(\Omega\)Y, P and Q so that the expression is PAQ. Replace the expression with PXX...XX0YY...YYQ with n X's and n Y's for the nth perm of the fundemental sequence. Category:Blog posts